Dependencies

A prime number \(p\) is called regular if it does not divide the class number of \(\mathbb {Q}(\zeta _p)\).

Let \(A,K\) be commutative rings with \(K\) and \(A\)-algebra. let \(B=\{ b_1,\dots ,b_n\} \) be a set of elements in \(K\). The discriminant of \(B\) is defined as

Let \(p\) be an odd regular prime. Then

has no solutions with \(x,y,z \in \mathbb {Z}\) and \(xyz \ne 0\).

Let \(K/F\) be a Galois extension of number fields whose Galois group \(\operatorname{Gal}(K/F)\) is cyclic with generator \(\sigma \). If \(\alpha \in K\) is such that \(N_{K/F}(\alpha )=1\), then

for some \(\beta \in \mathcal{O}_K\).

Let \(u \in F\) with \(F=\mathbb {Q}(\zeta _p)\) be a unit such that

for some \(a \in \mathcal{O}_F\). The either \(u = \epsilon ^p\) for some \(\epsilon \in \mathcal{O}_F^\times \) or \(p\) divides the class number of \(F\).

Let \(K\) be a number field and \(B,B'\) bases for \(K/\mathbb {Q}\). If \(P\) denotes the change of basis matrix, then

Let \(\zeta _p\) be a \(p\)-th root of unity for \(p\) an odd prime, let \(\lambda _p=1-\zeta _p\) and \(K=\mathbb {Q}(\zeta _p)\). Then

Let \(p\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \( \alpha \in \mathbb {Z}[\zeta _p]\) with \(\alpha =\sum _i a_i \zeta _p^i\). Let us suppose that there is \(i\) such that \(a_i = 0\). If \(n\) is an integer that divides \(\alpha \) in \(\mathbb {Z}[\zeta _p]\), then \(n\) divides each \(a_i\).

Let \(K\) be a number fields and \(\alpha \in K\). Then there exists a \(n \in \mathbb {Z}\backslash \{ 0\} \) such that \(n \alpha \) is an algebraic integer.

Let \(p\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \(\alpha \in \mathbb {Z}[\zeta _p]\). Then \(\alpha ^p\) is congruent to an integer modulo \(p\).

Let \(p \geq 3\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \( \alpha \in \mathbb {Z}[\zeta _p]\). Let \(x\) and \(y\) be integers such that \(x+y\zeta _p^i=u \alpha ^p\) with \(u \in \mathbb {Z}[\zeta _p]^\times \) and \(\alpha \in \mathbb {Z}[\zeta _p]\). Then there is an integer \(k\) such that

Let \(p \geq 5\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \(x, y \in \mathbb {Z}\) coprime.

For \(i \neq j\) with \(i,j \in {0,\dots ,p-1}\) we can write

with \(u\) a unit in \(\mathbb {Z}[\zeta _p]\). From this it follows that the ideals

are pairwise coprime.

Let \(p\) be a regular odd prime, \(x,y,z,\epsilon \in \mathbb {Z}[\zeta _p]\), \(\epsilon \) a unit, and \(n \in \mathbb {Z}_{\geq 1}\). Assume \(x,y,z\) are coprime to \((1-\zeta _p)\) and that \(x^p+y^p+\epsilon (1-\zeta _p)^{pn}z^p=0\). Then each of \((x+\zeta _p^i y)\) is divisible by \((1-\zeta _p)\) and there is a unique \(i_0\) such that \((x+\zeta _p^{i_0})\) is divisible by \((1-\zeta _p)^2\).

Let \(p\) be a regular odd prime, \(x,y,z,\epsilon \in \mathbb {Z}[\zeta _p]\), \(\epsilon \) a unit, and \(n \in \mathbb {Z}_{\geq 1}\). Assume \(x,y,z\) are coprime to \((1-\zeta _p)\), \(x^p+y^p+\epsilon (1-\zeta _p)^{pn}z^p=0\) and \(x+y\) is divisible by \(\mathfrak {p}^2\) and \(x+ \zeta _p^ky\) is only divisible by \(\mathfrak {p}=(1-\zeta _p)\) (for \(0 {\lt} k \leq p-1\)). Let \(\mathfrak {m}= \mathrm{gcd}((x),(y))\). Then:

1. We can write

   \[ (x+y)=\mathfrak {p}^{p(m-1)+1}\mathfrak {m}\mathfrak {c}_0 \]

   and

   \[ (x+\zeta _p^k y)=\mathfrak {p}\mathfrak {m}\mathfrak {c}_k \]

   (for \(0 {\lt} k \leq p-1\)) where \(m=n(p-1)\) and with \(\mathfrak {c}_i\) pairwise coprime.

2. Each \(\mathfrak {c}_k=\mathfrak {a}_k^p\) and \(\mathfrak {a}_k\mathfrak {a}_0^{-1}\) is principal (as a fractional ideal).

Let \(K/F\) be a Galois extension of \(F=\mathbb {Q}(\zeta _p)\) with Galois group \(\operatorname{Gal}(K/F)\) cyclic with generator \(\sigma \). Then there exists a unit \(\eta \in \mathcal{O}_K\) such that \(N_{K/F}(\eta )=1\) but does not have the form \(\epsilon /\sigma (\epsilon )\) for any unit \(\epsilon \in \mathcal{O}_K\).

Let \(L/K\) be an extension of fields and let \(B=\{ b_1,\dots ,b_n\} \) be a \(K\)-basis of \(L\). Then \(\Delta (B) \neq 0\).

Let \(F=\mathbb {Q}(\zeta _p)\) with \(\zeta _p\) a primitive \(p\)-th root of unity. If \(\xi \in \mathcal{O}_F\) and coprime to \(\lambda _p:=1-\zeta _p\) and if

for some \(\alpha _0 \in \mathcal{O}_{F_\mathfrak {p}}^\times \) (the ring of integers of the completion of \(F\) at \(\lambda _p\)), then the ideal \((\lambda _p)\) is unramified in \(K/F\) where \(K=F(\sqrt[p]{\xi })\).

Let \(p\) be a prime, \(K=\mathbb {Q}(\zeta _p)\) \(\alpha \in K\) such that there exists \(n \in \mathbb {N}\) such that \(\alpha ^n=1\), then \(\alpha =\pm \zeta _p^k\) for some \(k\).

Let \(\alpha \) be an algebraic integer all of whose conjugates have absolute value one. Then \(\alpha \) is a root of unity.

Let \(K\) be a number field, \(\alpha \in K\) and let \(\sigma _i\) be the embeddings of \(K\) into \(\mathbb {C}\). Then

.

Let \(K\) be a number field, \(\alpha \in K\) and let \(\sigma _i\) be the embeddings of \(K\) into \(\mathbb {C}\). Then

For \(n\) any integer, \(\Phi _n\) (the \(n\)-th cyclotomic polynomial) is a polynomial of degree \(\varphi (n)\) (where \(\varphi \) is Euler’s Totient function).

For \(n\) any integer, \(\Phi _n\) (the \(n\)-th cyclotomic polynomial) is an irreducible polynomial .

Let \(f\) be a monic irreducible polynomial over a number field \(K\) and let \(\alpha \) be one of its roots in \(\mathbb {C}\). Then

where the product is over the roots of \(f\) different from \(\alpha \).

Let \(K = \mathbb {Q}(\alpha )\) be a number field, where \(\alpha \) is an algebraic integer. Let \(B = \{ 1, \alpha , \ldots , \alpha ^{[K : \mathbb {Q}] - 1} \} \) be the basis given by \(\alpha \) and let \(x \in \mathcal{O}_K\). Then \(\Delta (B)x \in \mathbb {Z}[\alpha ]\).

Let \(K\) be a number field and \(B=\{ 1,\alpha ,\alpha ^2,\dots ,\alpha ^{n-1}\} \) for some \(\alpha \in K\). Then

where \(\sigma _i\) are the embeddings of \(K \) into \(\mathbb {C}\). Here \(\Delta (B)\) denotes the discriminant.

Let \(K\) be a number field with basis \(B=\{ b_1,\dots ,b_n\} \) and let \(\sigma _1,\dots ,\sigma _n\) be the embeddings of \(K\) into \(\mathbb {C}\). Now let \(M\) be the matrix

Then

Let \(K = \mathbb {Q}(\alpha )\) be a number field, where \(\alpha \) is an algebraic integer with minimal polynomial that is Eisenstein at \(p\). Let \(x \in \mathcal{O}_K\) such that \(p^n x \in \mathbb {Z}[\alpha ]\) for some \(n\). Then \(x \in \mathbb {Z}[\alpha ]\).

Let \(R\) be a Dedekind domain, \(p\) a prime and \(\mathfrak {a},\mathfrak {b},\mathfrak {c}\) ideals such that

and suppose \(\mathfrak {a},\mathfrak {b}\) are coprime. Then there exist ideals \(\mathfrak {e},\mathfrak {d}\) such that

Let \(K\) be a number field and \(B=\{ b_1,\dots ,b_n\} \) be elements in \(\mathcal{O}_K\), then \(\Delta (B) \in \mathbb {Z}\).

Let \(p \geq 3\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \(K=\mathbb {Q}(\zeta _p)\). Assume that we have \(x,y,z \in \mathbb {Z}\) with \(\gcd (xyz,p)=1\) and such that

This is the so called "case I". To prove Fermat’s last theorem, we may assume that:

• \(p \geq 5\);

• \(x,y,z\) are pairwise coprime;

• \(x \not\equiv y \mod p\).

If \(K\) is a number field and \(\alpha \in \mathcal{O}_K\) then \(N_{K/\mathbb {Q}}(\alpha )\) is in \(\mathbb {Z}\).

Let \(K=\mathbb {Q}(\alpha )\) be a number field with \(n=[K:\mathbb {Q}]\) and let \(B=\{ 1,\alpha ,\alpha ^2,\dots ,\alpha ^{n-1}\} \). Then

where \(m_\alpha '\) is the derivative of \(m_\alpha (x)\) (which we recall denotes the minimal polynomial of \(\alpha \)).

If \(K\) is a number field and \(\alpha \in \mathcal{O}_K\) then \(\operatorname {Tr}_{K/\mathbb {Q}}(\alpha )\) is in \(\mathbb {Z}\).

Any unit \(u\) in \(\mathbb {Z}[\zeta _p]\) can be written in the form \(\beta \zeta _p^k \) with \(k\) an integer and \(\beta \in \mathbb {R}\).

Let \(p\) be an odd prime, \(\zeta _p\) a primitive \(p\)-th root of unity and let \(K=\mathbb {Q}(\zeta _p)\). Then for any \(i,j \in {0,\dots ,p-1}\) with \(i \ne j\), there exists a unit \(u \in \mathcal{O}_K^\times \) such that \(\zeta _p^i-\zeta _p^j = u (\zeta _p-1)\).

Let \(p\) be an odd regular prime. Then

has no solutions with \(x,y,z \in \mathbb {Z}\) and \(\gcd (xyz,p)=1\).

Let \(p\) be an odd regular prime. Then

has no solutions with \(x,y,z \in \mathbb {Z}\) and \(p | xyz\).

Let \(\zeta _p\) be a \(p\)-th root of unity for \(p\) an odd prime, let \(\lambda _p=1-\zeta _p\) and \(K=\mathbb {Q}(\zeta _p)\). Then \(\mathcal{O}_K=\mathbb {Z}[\zeta _p]=\mathbb {Z}[\lambda _p]\).

Let \(p\) be a regular odd prime, \(\epsilon \in \mathbb {Z}[\zeta _p]^\times \) and \(n \in \mathbb {Z}_{\geq 1}\). Then the equation \(x^p+y^p+\epsilon (1-\zeta _p)^{pn}z^p=0\) has no solutions with \(x,y,z \in \mathbb {Z}[\zeta _p]\), all non-zero and \(xyz\) coprime to \((1-\zeta _p)\).

Let \(p\) be a regular prime and let \(u \in \mathbb {Z}[\zeta _p]^\times \). If \(u \equiv a \mod p\) for some \(a \in \mathbb {Z}\), then there exists \(v \in \mathbb {Z}[\zeta _p]^\times \) such that \(u=v^p\).